Let $A \in \mathcal{M}_{n\times n} (\mathbb{R})$ such that $A^2=\alpha A$ for some $\alpha \neq 0$.
Under this assumption, we have by induction that $A^{n}=\alpha^{n-1}A$; $n \geq2$; then:
$$\exp(tA)=I+ tA+\frac{t^2A^2}{2!}+ \dots \ =I+At+\frac{A\alpha t^2}{2!} + \frac{A\alpha^2 t^3}{3!} + \dots = \\ = I+ \frac{1}{\alpha}A(1+ \alpha t+ \frac{\alpha^2 t^2}{2!} +\dots)- \frac{A}{\alpha}=I+\frac{1}{\alpha}A\exp({\alpha t})- \frac{A}{\alpha}=I+A(\frac{\exp(\alpha t)-1}{\alpha}) $$
Are these steps correct?
COMMENT.- It is known that the function $x\to e^x$ operates over the algebra $\mathcal{M}_{n\times n} (\mathbb{R})$ (I don't know how the french word opère is translated in the English of Functional Analysis World). In simplest English, if $A\in\mathcal{M}_{n\times n} (\mathbb{R})$, the matrix $e^A\in\mathcal{M}_{n\times n}(\mathbb R)$
Now $\exp(tA)=I+A\left(\dfrac{\exp(a t)-1}{a}\right)$ is true each time that $A=aI$ in whose case $\exp(tA)=\exp(at)I$.
Is it possible that $A\ne aI$, say $A=aI+B$ for some non-zero $B\in\mathcal{M}_{n\times n}(\mathbb R)$?. From $A^2=aA$ we can get $A=aI$ only if $A$ is not a divisor of zero.
I tend to believe with @Somos above that the equality is true but I rather feel more comfortable not answering the question for now.